CIV E 530 - OPEN-CHANNEL HYDRAULICS SPRING 2005 - HOMEWORK 11 - SOLUTION

PROBLEM 1

∂Q/∂x + ∂A/∂t = 0

∂Q/∂x + B ∂y/∂t = 0

r = ∂y/∂t

∂Q/∂x + B r = 0

Q_d - Q_u = - B (Δx) r

Q_u = Q_d + B (Δx) r

Q_u = 650 + 280 (25,000) (4 mm/hr / 3600 s/hr) (1 m / 1000 mm)

Q_u = 657.78 m³   ANSWER.

PROBLEM 2

∂Q/∂x + ∂A/∂t = 0

∂Q/∂x + B ∂y/∂t = 0

r = ∂y/∂t

∂Q/∂x + B r = 0

r_a = 0.5 (r_u + r_d) = 6.0

Q_d - Q_u = - B (Δx) r_a

Q_u = Q_d + B (Δx) r_a

Q_u = 650 + 280 (25,000) (6.0 mm/hr / 3600 s/hr) (1 m / 1000 mm)

Q_u = 661.67 m³   ANSWER.

The second estimate is more accurate, because it is a central finite difference.

PROBLEM 3

F² = q²/(gy³)

y = [q²/(gF² ]^1/3 = 3.44 m.

F = v/(gy)^1/2 = 0.2

v = 0.2 (gy)^1/2

c_r= (gy)^1/2 = (9.81 × 3.44) ^1/2 = 5.81 m/s

v = 0.2 × 5.81 = 1.16 m/s

c_d = v + c_r = 1.16 + 5.81 = 6.97 m/s   ANSWER.

c_u = v - c_r = 1.16 - 5.81 = -4.65 m/s   ANSWER.

PROBLEM 4

Peak flood Q_p= 600 m³/s;

baseflow Q_b= 100 m³/s;

S_o = 0.0007;

A_p = 350 m²;

T_p = 85 m;

β = 1.65;

Δx = 9.8 km = 9,800 m;

Δt = 1 hour = 3,600 s.

The mean velocity V_p= Q_p/A_p = 600/350 = 1.714 m/s.

The wave celerity is: c = β V_p = 1.65 × 1.714 = 2.828 m/s.

The flow per unit width q_o = Q_p/T_p = 600/85 = 7.059 m²/s.

The Courant number C = c Δt/Δx = 2.828 m/s × 3600 s / 9,800 m = 1.039

The cell Reynolds number D = q_o/(S_oc Δx) = (7.059) / (0.0007 × 2.828 × 9,800) = 0.364

The routing coefficients are: C₀ = 0.168; C₁ = 0.697; and C₂ = 0.135.

The routing calculations are shown below.

Time	Inflows	C0I2	C1I1	C2O1	O2
0	100	-	-	-	100.000
1	130	21.795	69.721	13.514	105.030
2	150	25.148	90.637	14.193	129.979
3	180	30.178	104.581	17.565	152.324
4	220	36.884	125.497	20.585	182.966
5	250	41.914	153.386	24.726	220.025
6	300	50.296	174.302	29.734	254.332
7	360	60.356	209.162	34.370	303.888
8	450	75.445	250.995	41.067	367.506
9	520	87.180	313.744	49.664	450.588
10	600	100.593	362.548	60.891	524.032
11	550	92.210	418.325	70.816	581.351
12	490	82.151	383.464	78.562	544.177
13	370	62.032	341.632	73.539	477.203

The peak outflow is 581.351 and it occurs at time 11 hr.

The wave has diffused from a peak of 600 to 581, and has translated from t = 10 hr to t = 11 hr.   ANSWER.

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